用$g(x)=A*x^2+B*x+C$來代替原函式$f(x)$，設$g(x)$原函式為$G(x)$，顯然$G(x)=\frac{1}{3}*A*x^3+\frac{1}{2}*B*x^2+C*x$

$\int_a^b f(x) dx \approx \int_a^b g(x) dx=G(b)-G(a)$

代入化簡得

$$\int_a^b f(x) dx \approx \frac{a-b}{6}*[g(a)+g(b)+4*g(\frac{a+b}{2})]$$

二分割槽間，然後計算區間積分，通過期望容差來控制二分

如果$|S(a,mid)+S(mid,b)-S(a,b)|<15*\varepsilon$，則中止二分，並且認為

$$S(a,b)=S(a,m)+S(m,b)+\frac{S(a,mid)+S(mid,b)-S(a,b)}{15}$$

否則繼續二分

其中$mid=\frac{a+b}{2},\varepsilon為期望容差$

參考部落格：https://www.cnblogs.com/chy-2003/p/11644112.html

例題：

hdu 1724

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cmath>

using
 
 namespace std;

int T;
double a, b, l, r;

inline double sqr(double x)
{
    return x * x;
}

double f(double x)
{
    double t = sqr(b) - sqr(b) / sqr(a) * sqr(x);
    return 2 * sqrt(t);
}

double calc(double l, double r)
{
    double mid = (l + r) / 2;
    return (f(l) + 4 * f(mid) + f(r)) * (r - l) / 6
 
;
}

double simpson(double l, double r, double eps)
{
    double mid = (l + r) / 2;
    double st = calc(l, r), sl = calc(l, mid), sr = calc(mid, r);
    if (fabs(sl + sr - st) <= 15 * eps) return sl + sr + (sl + sr - st) / 15;
    return simpson(l, mid, eps / 2) + simpson(mid, r, eps / 2);
}

int main()
{
    // freopen("in.txt", "r", stdin);
    // freopen("out.txt", "w", stdout);
    scanf("%d", &T);
    while (T--) {
        scanf("%lf%lf%lf%lf", &a, &b, &l, &r);
        printf("%.3lf\n", simpson(l, r, 1e-6));
    }
    return 0;
}